How much heat is required to vaporize 33.8 g of water at 100 ∘c?

The heat required to vaporize 33.8 g of water at  is .

Further explanation:

Properties are categorized into two types:

1. Intensive properties:

These properties depend on the nature of the substance and not on the size of the system. If the system is further divided into a number of subsystems, the values of intensive properties remain unchanged. Temperature, refractive index, molarity, concentration, pressure, and density are some of the examples of intensive properties.

2. Extensive properties:

These properties depend on the amount of the substance. These are additive in nature if a single system is divided into several subsystems. Mass, heat, enthalpy, volume, energy, size, weight, and length are some of the examples of extensive properties.

Heat of vaporization:

It is the amount of heat that is needed to convert a unit mass of the liquid into its vapors. It is also known as the enthalpy of vaporization or heat of evaporation. It is represented by . Its unit is kJ/mol or J/mol.

The formula to calculate the moles of water is as follows:

                                                …… (1)

The given mass of water is 33.8 g.

The molar mass of water is 18.01 g/mol.

Substitute these values in equation (1).

The heat of vaporization of water at  is 40.7 kJ/mol. This implies 40.7 kJ of energy is required to convert 1 mole of water into vapors. So the heat required to convert 1.876 moles of water is calculated as follows:

Therefore the heat required to vaporize 33.8 g of water at  is 76.3532 kJ.

Learn more:

1. Calculate the enthalpy change using Hess’s Law:

2. Find the enthalpy of decomposition of 1 mole of MgO:

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Thermodynamics

Keywords: heat of vaporization, enthalpy of vaporization, heat of evaporation, 33.8 g, water, 40.7 kJ/mol, 76.3532 kJ, heat, intensive, extensive, molar mass of water, 18.01 g/mol, 1.876 mol.